help-49

[rain wasp]

any idea why my final expression is incorrect?

i have a minterm 00x, so ~x1~x2, as well as minterm xx1, so x3

but the expression doesn't match the truth table

[lyric charm]

x3 would give you 1 on input of 101

[rain wasp]

yeah, not supposed to happen

[lyric charm]

so you have an extra 1 in the bottom right of your kmap

that horizontal group of 3 doesn't give one minterm

[rain wasp]

what do you mean?

[lyric charm]

uh

basically this is how i would do it

[rain wasp]

why isn't the group on the second row maximized to 1x3?

[rain wasp]

i think you meant bottom left, which was why i got confused

[lyric charm]

what do you mean maximized

0 0 0 0
1 1 1 0

tell me what minterm gives this map

[lyric charm]

no, bottom right.

[rain wasp]

huh

[rain wasp]

x3

since the first two terms are changing, and the third term is stable

and since it's stable as 1, so it's x3

actually what

if x3 is 1 then the rectangle should be 1x3, not 1x4

so i need to break it down into two rectangles...?

oh

Minterm groups must be rectangular and must have an area that is a power of two (i.e., 1, 2, 4, 8...)

[lyric charm]

incorrect

x3 is

0 0 0 0
1 1 1 1 <-
      ^

this 1 is the source of errors

[rain wasp]

yes

if the area of one minterm should be a power of two, that means each side dimension should also be power of twos

and clearly 3x1 doesn't fit the bill

anyway thanks

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[inland patio]

Consider the Prokhorov metric: $$\tilde{\rho}(\mathbb{P},\mathbb{Q}) = \inf\bigl{ \delta>0 : \mathbb{P}(A) \le \mathbb{Q}(A^\delta) + \delta ;\text{and}; \mathbb{Q}(A) \le \mathbb{P}(A^\delta) + \delta ;\text{for all Borel } A \bigr}.$$ In the above theorem, what do they mean by a diagonal argument?

[grand pond]

psie

[inland patio]

$\tilde{\rho}(\mathbb{P}_n,\mathbb{P})\to0$ is saying that $\mathbb{P}_n(A) \le \mathbb{P}(A^{\epsilon}) + \epsilon$ for all $\epsilon>0$ and $n$ sufficiently large (and all Borel $A$). The fact that we have an $n$ dependence here drives me a little nutty.

[grand pond]

psie

[visual tiger]

Basically you have a kind of "double limit"

[visual tiger]

You have, in parallel to this, that $\mu(A^\epsilon)\to \mu(A)$ as $\epsilon \to 0$

[grand pond]

Rafilouyear2026

[inland patio]

Indeed, continuity from above.

[visual tiger]

From this, you can already write $\limsup_n \mu_n(A) \leq \mu(A^\epsilon) +\epsilon$ for all $\epsilon > 0$

[grand pond]

Rafilouyear2026

[visual tiger]

So you can take the limit of the right hand side as epsilon -> 0

[inland patio]

Hmm, ok. I thought n has an epsilon dependence, but maybe the order in which you take the operations makes this dependence irrelevant.

The inequality I posted only holds for $n>N_\epsilon$.

[grand pond]

psie

[visual tiger]

Yeah, but once you take the limsup, there is no dependence on epsilon anymore

If you start behaving like [good property] after some rank

Nevermind that the rank depends on epsilon

The limsup is the biggest behavior at "infinity"

So it also takes the good property

And removes the dependence in epsilon

[inland patio]

Ah, ok. Makes sense. Thank you.

[visual tiger]

As for the original argument, I can only think it's lacking some subsequence on the index of mu

[inland patio]

Hmm, yeah.

[visual tiger]

$\mu_{\phi(n)}(A) \leq \mu(A^{\epsilon_n}) + \epsilon_n$

[grand pond]

Rafilouyear2026

[visual tiger]

That's easier, take any sequence epsilon_n that decreases to 0, like 1/n

Then phi(n) is gonna be your "N(epsilon_n)", so your N(n)

[inland patio]

Ok.

[visual tiger]

Well you need to failproof it a little bit by adding that phi(n+1) can't be smaller than phi(n)

So phi(n+1) = max(phi(n)+1, N(n+1))

And we're done

Since there's a subsequence of mu_n that follows that pattern

Mmmh, wait

Ok no I've been taking it the wrong way around

You might be able to find some epsilon_n that works

If you take $\epsilon_n = \inf{\epsilon > 0: :\forall k\geq n, \mu_k(A) \leq \mu(A^\epsilon) + \epsilon}$

[grand pond]

Rafilouyear2026

[visual tiger]

The set is not empty because epsilon = 1 is in it

It is non-increasing

So converges, and if it converges to epsilon > 0

Then applying the property to epsilon/2 yields a contradiction

So it converges to 0

[inland patio]

But your argument works as well, right? I.e. taking limsup first and then epsilon to 0.

[visual tiger]

Yes

Funnily enough, I'm in a "limit theorems" class right now

[inland patio]

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[median holly]

reupload cuz my wifi was gone

[median holly]

one of the sol is cosx=-1/2

trying to solve this for 4hrs

[dreamy lichen]

whats ur current progress?

[median holly]

nothing

no clue

[dreamy lichen]

2-2cos^2 is sin^2

that'd remove the +2

then I'd try to expand sin(2x) and cos(4x)

and then it'll hopefully be factorable

hmm

just tried putting it to wolfram alpha

you'd need to solve a 7th degree polynomial

[median holly]

damn

omfg

[dreamy lichen]

seems like it doesnt have nice sols

[median holly]

my math teacher said we can factorize it

if i was mistaken one of the sol is cosx= -1/2 or 1/2

but still

idk how

[dreamy lichen]

maybe i miscopied sth

my input seems correct

unless these are multiplications

which they shouldnt be but idk

[median holly]

well no multiplication

[dreamy lichen]

then the question is wrong ig

[surreal charm]

no general solutions

[median holly]

x= pi/6 still a solution

you all can try

[surreal charm]

hm..

[dreamy lichen]

doesnt look like it

[surreal charm]

the sqrt3 says transform to sin(x+pi/6)

[median holly]

already tried

[surreal charm]

!show

[midnight plank]

Show your work, and if possible, explain where you are stuck.

[dreamy lichen]

yeah, pi/6 isnt a sol

all the solutions are equally ugly

[median holly]

oh

[surreal charm]

yh

most simplified
$2\sin(\frac{\pi}{6}+x)+2\sin(2x-\frac{\pi}{6})-4\cos(4x)\cos(x)+1=0$

[grand pond]

scoob

[surreal charm]

ask

[median holly]

he is sleeping bruh

[surreal charm]

leave it

[median holly]

my class gonna start tmr i will ask him

[midnight plank]

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[woeful turret]

how to integrate this

like how to start

[fickle sierra]

What algebraic manipulations have you done?

[woeful turret]

i mean i tried taking x out of numerator

thats it

[bold peak]

Partial fractions maybe?

Yeah partial fractions works I'm 70% sure

[deep vine]

Doesn't partial fraction only work for rational functions

[bold peak]

I mean the idea works for anything

Separate the denominator and see where that gets you

[woeful turret]

huhh

i didnt think of that

alr then ty

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[slow thorn]

haha this one was funny

[woeful turret]

what is funny about it sir

[slow thorn]

i didnt expect the partial fractions when i started

[woeful turret]

bro but it was kinda funny cause i got the answer without integrating

[slow thorn]

yeahhh

[woeful turret]

yea lmao

[fickle sierra]

bruh

[lusty python]

anymore proofs to show us lol

[fickle sierra]

their cooking

[slow thorn]

wait helpful???

wow cgats

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[obsidian glen]

how do I show equivalence between (0,1) and [0,1]

[dreamy lichen]

Are you trying to show they have the same cardinality

Do you need to construct an explicit bijection or can u use e.g. cantor schroder berstein and only construct an injection both ways

[obsidian glen]

Yes I think, I just need to show the existence of one and not explicitly construct one

[dreamy lichen]

and do you know cantor schroder berstein theorem? It allows you to prove that 2 sets A,B are of the same cardinality just by constructing 2 injections A->B and B->A

[obsidian glen]

yes I know that

[dreamy lichen]

okay, great

try doing that then. In one direction it should be trivial

[obsidian glen]

I need the injective map [0,1] \to (0,1) because the other direction i can very well use the identity map

[dreamy lichen]

try making the map to first shrink the interval to some smaller one

and then shift that interval so that it fits inside (0,1)

would u know how to make an injective map to sth like [0, 1/2]?

[obsidian glen]

yeah

1-x/2

[dreamy lichen]

that'd be to [1/2, 1]

but yeah, why not

now just shift it to fit inside (0,1) and youre done

[obsidian glen]

3/4-x/2 or smth

should do

[dreamy lichen]

yeah, or 1/4 + x/2

[obsidian glen]

hmm

[dreamy lichen]

there is also a way to make an explicit bijection (there is an analogy with hilberts hotel), but this double injection is simpler

[obsidian glen]

tell me the analogy as well

[dreamy lichen]

we first start by sending everyone from (0, 1) to their room number

but since we want [0,1] -> (0,1), there are 2 "guests" left

0 and 1

so we can send 0 e.g. to room 1/2

but then we need to send guest 1/2 elsewhere

so we send him to 1/4

then we need to send 1/4 elsewhere

so we send him to 1/8

...

and every guest finds their room, and 0 now has a room

you can similarly send 1 to 3/4, 3/4 to 7/8, ...

and therefore accomodate the second guest as well

[obsidian glen]

makes sense

[dreamy lichen]

you can basically pick a countable sub-hotel of the uncountable (0,1) hotel and that countable sub-hotel behaves exactly like hilberts hotel

[obsidian glen]

I get it

ty !!

[dreamy lichen]

np

[obsidian glen]

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[ruby latch]

I'm trying to solve a problem that's either optimization or curve fitting.

A user comes to me with a list of ratings (from 0 to 100%) they've given to a bunch of books. Each book has several themes, each with a percentage (e.g. Sci-fi 22%, Drama 20%, Mystery 19%, Comedy 16%, Romance 15%, Crime 8%). These themes are not chosen by the user.

My goal is to return to them a list of their preferences for each theme from -10 to 10.

The way the preferences work is that each book has a base rating (from 0 to 100%, not given by the user) and this is multiplied by 1.02^total preference, where total preference for a book is the sum of each theme's preference multiplied by how much that theme applies to the book.

One problem is that the rating resulting from this is not similar to the rating given by a user, and so the resulting batch would need to go through normalization before being compared to the user ratings.

At the end of that, I'm trying to find a way to minimize the (square) differences between the ratings given by the calculated preferences and the ratings given by the user.
Is there any sort of algorithm that works for that?

[midnight plank]

@ruby latch Has your question been resolved?

[ruby latch]

For curve fitting problems, the best lead I could find was the Gauss-Newton method, but that seems to be for finding a function f for y=f(x), where x and y are real numbers.
But for me, only y is a real number, and x is a vector, I guess? I'm not sure if x has to be a real number.
But what seems like a larger problem is that I don't think my function is differentiable, so the method doesn't work?

Since, sure, for simplicity let's say there's 3 themes.
The function would be y = r * 1.02^(P_1 * T_1 + P_2 * T_2 + P_3 * T_3), and like maybe finding P_1, P_2, and P_3 would be similar to finding a, b, and c in y = ax^2 + bx + c.
But then the normalization step, which is very necessary, is no longer a function of a single book and preferences.

I guess I could always brute force it by looping through each preference and calculating the MSE for each of the 21 values it could be in until no more improvements can be made.

That would make the (probably false) assumption that each preference has an independent effect on the result, but I imagine it wouldn't be too far off the mark.
I could even optimize in groups of 4 or 5, which would probably finish in time.

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@ruby latch Has your question been resolved?

[ruby latch]

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[strange rock]

I need help on this

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[vivid yoke]

A silly question, how would you find the 40th decimal place of 11/57

[vivid yoke]

W/o a calculator

[balmy turret]

Find the period in the decimal composition by making the euclidean division of 11 * 10^n by 57 repeatedly, and then apply for the right number of periods

[vivid yoke]

hmm

[rich yew]

Probably first find the period where it repeats, then repeat that period as many times as fits into 40 (GCD) and then count in the remaining period?

Not the most effecient.

[vivid yoke]

Yeah I see, I just have no idea how to find the period

I rely too much on calculator I have no idea how to do division now

[balmy turret]

Idk the carracterisation of it

It just is <57

[vivid yoke]

so I just take a random n then do division until it repeat?

[balmy turret]

You start for n=0,1,... util it repeat itself

[runic hamlet]

if you do the division by hand like you did in fourth grade then you will realize that it is the same as computing 11*10^n mod 57

[vivid yoke]

[balmy turret]

• you want the quotient not the rest

[runic hamlet]

hence this is essentially the order of 10 mod 57

which you can compute in various way

[balmy turret]

This

[runic hamlet]

all somewhat painful

[balmy turret]

This

[runic hamlet]

phi(57)=36 so you would only need to test n that are divisors of 36

but there are so many of those

[vivid yoke]

hang on hang on gimme a min, I have never taken a proper number theory class these euclidean division and phi terms are kinda confusing to me

Gimme a min

[runic hamlet]

do this

like actually do it

at least three steps

of each

[vivid yoke]

OHHHH

I see

I just remembered how to do division with decimal

wait how would I know when it repeat

[runic hamlet]

when 11*10^n = 11 mod 57 again

[vivid yoke]

Let me try that

[woeful turret]

do it yourself 🫠 🫠

[runic hamlet]

(this is gonna take 18 steps, dont actually do it by hand pls)

do 11/63

[woeful turret]

ws joking

[vivid yoke]

Okay okay

oh wait

Does the phi thing

It means the period of smth when doing mod right?

OHHHH

I seee

I see how that works

Tyyy

[runic hamlet]

well more or less

its guaranteed to be a multiple of the period

due to general group theory

[vivid yoke]

Okay I have no idea what that is

I might have to work on my number theory skill

But ty

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[graceful drum]

When does $\frac{3^{11n}-17}{2^{11n}-17} \in \mathbb{N}$ for $n \in \mathbb{N}$.

[grand pond]

CherryMan

[graceful drum]

we found that n is even and were trying some size arguments

$S = \frac{3^{11n}-17}{2^{11n}-17}$ then $S - \left(\frac{3}{2}\right)^{11n}$ converges to 0. So for large enough n it is never an integer

[grand pond]

CherryMan

[graceful drum]

maybe

[signal ibex]

try taking the numerator and denominator

and doing (mod 3) to both

[graceful drum]

i already said this

n evne

[signal ibex]

what im trying to do is find a number that is a factor of the denominator but not the numerator

[graceful drum]

i tried that for 3,5,7,11

only found conditions on n

like n is not congruent to 2 mod 3 or not congruent to 9 mod 10

there has to be a better way

[signal ibex]

theres gotta be a trick

n even because otherwise the denominator is div by 3 but the numerator is never divisible by 3

let's just write down that n = 0 is a solution

just have that there

[graceful drum]

n is natural

oh wait

0 is a natural number right ha

[signal ibex]

now mod 5, n has to be 1 mod 5

wait

[graceful drum]

<@&268886789983436800>

[signal ibex]

my god this is annoying

[graceful drum]

right this shouldnt be so difficult

[signal ibex]

im going to try mod 10

[surreal charm]

-# side note original q was framed off

so there might be no nice way to prove it

[signal ibex]

the answer is probably never (or only n=0)

im trying to prove that

[surreal charm]

yh graph says

[signal ibex]

mod 15 maybe?

no that does nothing

[signal ibex]

mod 7 does something

GODDAMNIT

numerator also divisible by 7

oh my god

[graceful drum]

lel

[signal ibex]

tried mod 14, also did nothing

[graceful drum]

@signal ibex does this work or is it faulty

[signal ibex]

i dont think that works

n can't be 3 or 9 mod 9

mod 18, that leaves us with candidates

2, 4, 6, 8, 10, 14, 16

the move might be to test out the powers of 3

there's gotta be a better way

[fallow scarab]

since 11 is in the exponent, did you try mod 11

[signal ibex]

can't be 9 mod 11

[graceful drum]

how does that make sense

[fallow scarab]

$3^{11} \equiv ? \mod{11}$

[grand pond]

riemann

[graceful drum]

oops

yeah its 3

[fallow scarab]

3^n is surely easier to deal with than 3^(11n)

[signal ibex]

2^11 is 2 mod 11

and 17 is 6...

yeah this does the job

the denominator is always 0 mod 2 but the numerator never is

mod 22 i think is the test we need for the original problem

[graceful drum]

why is the denminator always 0

[signal ibex]

2^n - 6

[graceful drum]

then?

[signal ibex]

thats always even

[graceful drum]

bruh that means its 11k + (2^n -6)

[signal ibex]

right

[graceful drum]

we cant say for sure that its even or not right?

[signal ibex]

that is true

[midnight plank]

@graceful drum Has your question been resolved?

[midnight plank]

@graceful drum Has your question been resolved?

[spiral rock]

Isn't this good?

[graceful drum]

no cus the fractional part of (3/2)^n may be very small always

always smaller than how close the difference is

[spiral rock]

right

[lusty python]

ofc that it's when $2^{11n} - 17 \mid 3^{11n} - 17$

[grand pond]

1 divided by 0 equals Infinity

[lusty python]

just need to grind some info on this

[vivid swallow]

kek

[lusty python]

got this so far

[vivid swallow]

bro 😭

3rd line randomly adds 17*3^11n

how is bro fumbling so much

[signal ibex]

i think literally all of that screenshot is wrong

[signal ibex]

maybe we can let this be an integer

rearrange to get

then do modulo some number to find a contradiction

most likely 17

double check that picture

if this is correct, there are at most 289 combinations to check

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[untold dirge]

How can we show that set of all rational numbers can be written as like rigorous proof

[wanton spade]

,rccw

[supple sigil]

one direction of the inclusion should be trivial

[untold dirge]

What that means?

[molten mauve]

a common way to show two sets A and B are equal is to show A ⊆ B and A ⊇ B

plante is saying that one of these is trivial

[untold dirge]

OK then why we need to prove two set equal right here

[supple sigil]

Cause the question asks you to

[untold dirge]

actually I don't understand question right here also can you explain that

[molten mauve]

hm well what's confusing you the most

[untold dirge]

Hmm all of that means what we're trying do right here

[main current]

We've gotta have some idea what you know / don't know

[untold dirge]

We know that the set of all rational numbers can be written like this but how we can prove like rigorously and someone has told me that one of that is trivial (equality of sets) but how?

[lavish venture]

have you even done the first one yet btw?

i assume no

[lavish venture]

take arbitrary elements from each and show that they are necessarily in the other set

do this both ways and you establish equality

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[twilit field]

Am I tripping or it this just $\int_{0}^{1} \frac{(1-x)^{10}}{\sqrt{x}}dx$

[grand pond]

wai

[bold peak]

Believe so yeah

[twilit field]

cool, thanks

.close

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[shut canyon]

braids have analogies with basic algebra ?

[supple sigil]

braids?

braid group?

[shut canyon]

for example, 1+1, or a(b+c), multiplication, addition, axioms, small lemmas have analogues?

distributivity, and so on

for example have two parallel strings = is 0? and +1 is ⤬ ?

═ • ⤬ = ═ ?

⤬ + ⤬ = ⤬⤬ ?

What is ⤬•⤬

[midnight plank]

@shut canyon Has your question been resolved?

[supple sigil]

how are you defining multiplication and addition

braids have a natural composition by drawing them left to right and then "straightening" the strings

[midnight plank]

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[coral belfry]

,tex
Hi there. is the negation of the first statement correct?

\vspace{0.15cm}
$\lim_{x\to \infty} f(x) = l \implies \forall \epsilon >0 ,\exists M>0 : x>M \implies \lvert f(x) -l\rvert <\epsilon $

\vspace{0.24cm}
$\lim_{x\to\infty } f(x) \neq l \implies \exists \epsilon >0 \ \text{s.t.} \ \forall M>0 : x>M \ \text{and} \ \lvert f(x) -l\rvert >\epsilon $

[grand pond]

fijokazż

[supple sigil]

Missing a ∃ before x>M

[coral belfry]

oh yeah thatd make sense

okay thanks

.solved

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[tardy bloom]

.

Number of roots of the equation
1/(x+1)³ - 3x + sinx=0

[tardy bloom]

I differentiated the equation and got f'(x)<0 , so 1 root but the answer given is 2 roots

[slow socket]

와

[tardy bloom]

?

[vivid yoke]

Can you show your work?

[tardy bloom]

f'(x) = -3/(1+x)⁴-3-cosx <0

[vivid yoke]

Hm

well

f(x) is not continuous on R

If you compute lim f(x) when x approaches -1 from the right and from the left you will understand why

Also, make a rough sketch

[tardy bloom]

Bruh

f(-1+)= infinity
f(-1-)= -infinity

So two roots

Thanks @vivid yoke

[vivid yoke]

Yeah

,w plot 1/(x+1)^3 -3x + sin(x)

[grand pond]

Results provided by WolframAlpha

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Requested by yobi_fx.

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[thorny pelican]

Find the number of permutations of n different things taken r at a time such that 3 specific things occur together.

[thorny pelican]

i took out the combination to be (n-3) C (r-3) and then the ways these 3 objects can be arranged is 3!

im still missing a part

[olive matrix]

do you mean:
find the number of permutations of 52 different cards taken 5 at a time such that you have the ace, king, and queen of hearts?

[thorny pelican]

no the question is find the number of permutations of n distinct things taken r together in which that 3 particular things must occur together.

[vivid yoke]

For now you only have the number of combinations of r objects

We need number of permutations right?

[thorny pelican]

yes

[vivid yoke]

Wait

"Occur together" like they're next to each other ?

Is this what you meant?

Or those 3 objects just have to be in r objects we have chosen

[thorny pelican]

the question is given like this 😭

i think

we have to

[thorny pelican]

yeah this one

[scenic wyvern]

is there any other context? perhaps showing the original question would be better.

[visual tiger]

Are you sure "permutation" is the exact wording, and not something like "find the number of ways to put n different things in groups of r such that 3 specific things are in the same group"?

!xy

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[visual tiger]

ok, weird wording coming from the book but ok

and no idea if the order of the groups of r matters or not

[vivid yoke]

When in doubt, do both

[thorny pelican]

the solution i can send asw

if it helps

3! * (n-3) C (r-3) * (r-2)!

I did not understand how the (r-2)! came

[vivid yoke]

Oh they meant this

Those 3 objects have to go together, like O1-O2-O3

[visual tiger]

wdym

isn't it just "they're in the same pack of r elements"

[vivid yoke]

No? they multiply by (r-2)!

And also 3!

Ig 3! Is for the permutations of those 3 objects

And (r-2)! Is for the permutations of (r-2) objects ( treat those 3 specific objects as one )

[visual tiger]

So the wording is completely wrong

It should be "how many ways to arrange r elements picked from a set of n elements, such that 3 specific elements are together"

[visual tiger]

@thorny pelican do you understand the question better once written that way

[thorny pelican]

oh

[thorny pelican]

alright

thanks

wait

so why are we treating the 3 objects as one

is it because it is compulsory to take those 3

[visual tiger]

Those 3 are determined to be together

so once you've chosen the other r-3 objects to arrange

The arrangement has to be like [random object] [random object] .... [random object] [ordered group of 3 objects together] [random object] ...

so if you treat the ordered group of those 3 as a single thing

[thorny pelican]

ohh yeah

[visual tiger]

you have r-2 things to order

[thorny pelican]

alr thanks

.close

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[midnight plank]

Please don't occupy multiple help channels.

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[hybrid crow]

Any ideas on how to do this?

[hybrid crow]

also if I do (b) do I even need to bother doing (a)

[runic hamlet]

you could start with an example

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[fallow scarab]

<@&268886789983436800> spam

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[small warren]

Is my proof correct?

[small warren]

Maybe I should've added "through transversals" for the first reason, and I think that saying the measure of angle A was equal to itself was redundant because the last line right after it didn't need it to complete the proof, if my proof even answers the original question (which I'm not 100% sure if that's what it's asking for).

[gaunt jetty]

I feel like you stopped short of proving what rhe question is asking you for

Like tou are meant to show AD/DB = AE/EC right

[gaunt jetty]

Also yes it is redundant bcuz u only need two pairs

[small warren]

Yep I unpaused the video and that's what was being asked apparently, unfortunately I misinterpreted the question.

[small warren]

Right

Tyvm!

.close

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[hybrid crow]

Any hints?

the division algorithm feels useful here somehow

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[thin glade]

<@&268886789983436800> unsolicited philanthropy

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[flat geyser]

break into cases: when both are positive, what is ab? when both are negative, what is ab? when one is pos and one is neg, what is ab?

[outer copper]

?

[flat geyser]

for the first case, if a and b are positive, meaning $a>0$ and $b>0$, then $ab>0$ (this property might be in your notes or something like it), meaning $ab\neq0$

[grand pond]

JamR_71111

[outer copper]

but thats for ordered fields isnt it

this is just regular field

[flat geyser]

ah should have asked, let me see

Okay, I think I remember how to deal with this in a general field

[outer copper]

ty mods

[flat geyser]

split into cases 1) $a=0$ and 2) $a\neq0$

[grand pond]

JamR_71111

[flat geyser]

the $a=0$ case is simple but $a\neq0$ requires using the axiom about multiplicative inverses

[grand pond]

JamR_71111

[outer copper]

i dont think i can split into cases tho

bcz that's my implication

it would work if it said "if a = 0 or b = 0, then ab = 0"

[twilit jetty]

no matter which way youre doing this, you need to consider the nonzero stuff

[twilit jetty]

going this way would consider zeros only, you gotta think of positive/negative to go the other way

[flat geyser]

the set-up here is that for any given field, an element a in that field must be either 0 or not 0
if we split it into the two cases, what we want to show is that if a isn't 0, then b must be

[twilit jetty]

thats why this will work

[flat geyser]

they aren't working with an ordered field so the standard pos/neg thing doesnt work

[outer copper]

do i have to show that they are both 0?

[flat geyser]

all you have to show is that at least one of them has to be

that if one of them isn't, then the other is

do you want me to get you going a bit further than the vague set-up

[outer copper]

assume $a \neq 0$, then we can do $\frac{1}{a} \cdot (ab) \implies (\frac{1}{a} \cdot a) \cdot b \implies 1 \cdot b \implies b$.
from the question, we have $ab = 0$, so $\frac{1}{a} \cdot (ab) \implies \frac{1}{a} \cdot 0 \implies 0 = b$

[grand pond]

joseph

[outer copper]

is this ok?

[flat geyser]

a bit messy with the $\Rightarrow$'s, but yes that's exactly the standard way to show it

[grand pond]

JamR_71111

[outer copper]

okay, so do i have to show when a = 0?

or can i just leave it at that and say something like

WLOG lol

WLOG, assume $a \neq 0$

[grand pond]

joseph

[flat geyser]

you've already shown $a\neq0\Rightarrow b=0$ and certainly $a=0\Rightarrow a=0,$ so you've already done the only two possible cases $a=0$ or $a\neq0$

[grand pond]

JamR_71111

[outer copper]

ahh

[flat geyser]

dont even need a WLOG since it's all on that $a\neq0\Rightarrow b=0$

[outer copper]

okok

[grand pond]

JamR_71111

[outer copper]

thank you

[flat geyser]

Np

[outer copper]

.close

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[cerulean oyster]

This is essentiallly a definition

[storm sedge]

what im guessing is to find the instantaneous rate of change at -1.999... and at -2.001

is my guess

[cerulean oyster]

Althought it requires some testing

[cerulean oyster]

yea

[storm sedge]

bruh

thats so much wokr

i hate this

[cerulean oyster]

theres also the possibility of checking using the 2nd derivative

given this is a polynomial

[storm sedge]

this is advanced functions so we havent done derivatives yet

im taking calc next semester tho

[cerulean oyster]

oh yea

This process is simulating the 1st derivative of a function fyi

[storm sedge]

yuh i know like the basic stuff for derivatives but my teacher wants us to do it some other wack

way

its so annoying

[midnight plank]

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[tawny tiger]

this is the first option to a question. it has 4 options and asks which of the following is true, only one option is correct. all of them felt seemingly correct, it gave result to this question as well in the option. the option felt correct but its not correct, looking at the solutions gives me some hints to how limits questions are supposed to be proceeded.

[tawny tiger]

i would like to see how limits work

provide with a soln as in how you would proceed to solve this question

[lethal path]

what's the original question?

[tawny tiger]

the questions asks which of the following options are correct.

gimme a min

ill send

[lethal path]

alright, why do you think option A is correct?

oh that's so tricky, I see

[tawny tiger]

i think because the argument will tend to 1

[lethal path]

consider the domain of that function

[tawny tiger]

domain of cosec inverse

[lethal path]

the important thing is that as $x \to \infty$, $\frac{3x^2 - 4x + 2}{3x^2 + x + 7}$ approaches $1$ from below

[lethal path]

yes

[grand pond]

south

[tawny tiger]

yes

understood this

[lethal path]

therefore cosec inverse of this does not exist

cosec inverse of 1 exists, but 1 is never attained

[tawny tiger]

so from which side is the argument tending towards 1

[lethal path]

from below, so 0.9, 0.99, 0.999...

[tawny tiger]

and thats outside the domain

right

[lethal path]

exactly

[tawny tiger]

okay

likewise in the 2nd option

i think its the similar way

[lethal path]

yes, so the right-hand side of x = 1 does not exist

it's not in the domain

but the left-hand side exists (and it's continuous)

so yes, the limit exists as a one-sided limit

[tawny tiger]

ohh

ill try the 3rd and 4th options and get to you in a while

once again

[midnight plank]

@tawny tiger Has your question been resolved?

[tawny tiger]

Third option I think I am missing something not sure how to proceed

For the fourth option I am stuck here

[
\lim_{x \to inf}
\left[
x\left{
\left(1+\frac{2}{x}\right)^{\frac{1}{5}}

\left(1+\frac{1}{x}\right)^{\frac{1}{4}}
\right}
\right]
\left[
\frac{20 + \frac{1}{\pi x^5}}{3 + \frac{1}{x^5}}
\right]
]

[grand pond]

CriTiCABLE

[lethal path]

$(1 + h)^n \approx 1 + nh$ for $h \to 0$

[grand pond]

south

[tawny tiger]

okay

next what

[lethal path]

for the third option, you'll need $\frac{\sin x}{x} \approx 1 - \frac{x^2}{6}$

now how about with $x \mapsto 1/x$?

[grand pond]

south

[lethal path]

it's easy after you apply it

[tawny tiger]

what do we do with the last term

[lethal path]

you can just take it as 20/3

[tawny tiger]

hm right

i get the result as 1/10

is that right

[lethal path]

no

[tawny tiger]

sorry 1

[lethal path]

yeah 1, now that's correct

[tawny tiger]

i took 2/3 by mistake

[tawny tiger]

sure thanks

3rd one ill figure out

.close

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[lethal path]

third one is tricky, so okay use this fact

then what I mean is $\lim_{x \to 0} \frac{1}{x^2} - \frac{1}{\sin^2 x}$

[grand pond]

south

[lethal path]

no worries!

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[dreamy lichen]

Why does center of mass / centroid exist? (im okay with assuming uniform distribution). Center of mass apparently has this property that we can balance stuff on it, mathematically, that'd mean that if we draw any line through the COM, the torques (sums of distances to the line * mass) on both sides are equal.

Now why should a point which satisfies this property exist? If we took torque to be only dependent on mass, then the corresponding balancing statement would be that areas on both sides of any line passing through the COM are equal, and such a COM wouldnt exist for most shapes. So why does the normal COM exist? And why even is torque = distance * weight?

[jaunty ivy]

torque is not distance*mass

[dreamy lichen]

Oh, right, I should probably call it distance * weight then

my point is mainly why it should be proportional to distance

and not distance^2, or even no distance

i suppose that we could say that torque = distance * force by definition, but in that case, why should 0 total torque mean that the object wont rotate / tilt

[jaunty ivy]

it is like pushing this door. the door rotates about the left vertical axis (the hinge axis). if u take the moment about that axis and push the door on that same axis u get that the force vector is parallel to the displacement vector (we define the displacement vector as starting at any point on the hinge axis to ending at any point on the line of action by the force)

torque is the cross product r cross F

[dreamy lichen]

Hmm, i dont really see how this would imply that the torque is proportional to distance from axis of rotation. Why should the door accelerate (rotationally) twice as fast when i apply the force twice as far? Why not e.g. four times as fast?

[jaunty ivy]

i have a feeling that this comes down to experimental results and not a mathematical one

[small crypt]

cause its directly proportional to r vector

[small crypt]

yeah

[dreamy lichen]

hmm, that's possible

[small crypt]

but i think

[surreal charm]

https://scholarship.haverford.edu/cgi/viewcontent.cgi?referer=https://www.google.com/&httpsredir=1&article=1494&context=physics_facpubs

[small crypt]

isnt torque defined as -F cross r vecor

[surreal charm]

this may have context

[jaunty ivy]

yes, from what i understand that definition is motivated by experimental results

[small crypt]

makes sense

[dreamy lichen]

yeah, i get that, but why? Why should this definition correspond to how the object rotates? Why should 0 net torque imply it? Ig its gonna be just experimental

[dreamy lichen]

Alright, thanks, ill read it..

[jaunty ivy]

this sums up what i think as well

https://physics.stackexchange.com/questions/791239/why-does-torque-increase-with-radius-or-distance-from-the-centre

[small crypt]

cause only perpendicular forces to the r vector can produce any rot. effects

[dreamy lichen]

Oh, it can also follow from newton's second law

[small crypt]

o net torque implies-;
0 force
0 r vector
angle b/w the two vectors = 0

either of these 3

[dreamy lichen]

Alright, thanks everyone, this solves my second question

Why does center of mass / centroid exist? (im okay with assuming uniform distribution). Center of mass apparently has this property that we can balance stuff on it, mathematically, that'd mean that if we draw any line through the COM, the torques (sums of distances to the line * mass) on both sides are equal.

Now why should a point which satisfies this property exist? If we took "torque" to be only dependent on mass, then the corresponding balancing statement would be that areas on both sides of any line passing through the COM are equal, and such a COM wouldnt exist for most shapes. So why does the normal COM exist?

just reposting the first part of the question

[midnight plank]

@dreamy lichen Has your question been resolved?

[molten mauve]

i don't see why torque being only dependent on mass means the corresponding balancing statement might not have a solution

ah, you mean for all plumb lines

ok so this has to do with the fact that torque is linear

[dreamy lichen]

Oh, that actually kinda makes sense

is there any way to prove it using the fact that torque is linear?

i dont need a rigorous proof, im okay with some handwaving

[molten mauve]

Let $\mathbf{p}$ be a point. Let $\fn{\rho}{\bb{R}^3}{\bb{R}}$ be the density distribution of your object. Then net torque with respect to $\mathbf{p}$ is
\[ \tau_{\mathrm{net}} = \int \rho(\mathbf{q})(\mathbf{q}-\mathbf{p})\times\mathbf{g}\,d\mathbf{q} \]
and by linearity if you want $\tau_{\mathrm{net}} = 0$ you can distribute to get an equation for $\mathbf{p}$ which you can solve for:
\[ \int\rho(\mathbf{q})\mathbf{p}\,d\mathbf{q} = \int\rho(\mathbf{q})\mathbf{q}\,d\mathbf{q}. \]

[molten mauve]

g is gravity vector

[dreamy lichen]

What's p?

oh a point

im blind

Oh damn, i get it

i was focusing too much on balancing over a single line, but when we instead look at the torque as a vector and do all directions simultaneously, it becomes obvious by linearity

[dreamy lichen]

Tysm, it finally makes sense

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[molten mauve]

i guess you have to show that vector valued integrals are independent of choice of basis

[dreamy lichen]

I'm probably going to accept that as intuitive enough for now

[midnight plank]

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[floral ruin]

how do I figure out f(x) ?

[wispy mortar]

f(x) is y/x^2 right?

[floral ruin]

ohh

that was silly of me

thanks

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[woeful turret]

in terms of only x he was asking

[woeful turret]

not really u can reopen

[floral ruin]

.reopen

[midnight plank]

✅ Original question: #help-49 message

[floral ruin]

yea I was confused in that way

[woeful turret]

i would break it as
f(x)+f(1/x) = (x^2-2)+ (1/x^2 -2)

do u see what im getting at

[floral ruin]

just a sec

ohhh

got it sirr

[woeful turret]

no sir bruh im younger than u lmao but nw

[floral ruin]

ayo cmon are you really younger than me? what is your age?

[woeful turret]

17

[floral ruin]

same 😭

[woeful turret]

o then mb

[floral ruin]

tho thanks

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[balmy cypress]

Why |P(A)|=2^{n}?

[hexed barn]

the quick and easy explanation is because to form a subset of A, for each element in A, you choose whether to include it or not in the subset - 2 choices.
since A has n elements (presumably), that's 2^n different combinations of inclusion/exclusion to make, which is exactly the cardinality of the power set of A

[lyric charm]

what's n

[balmy cypress]

Elements of the set

A

[hexed barn]

you mean the number of elements in A?

[lyric charm]

^

[balmy cypress]

Yes

[lyric charm]

to make a subset of A you need to go through each element of A and answer the question "will it belong to my subset or not"

yes/no question => 2 options

2 * 2 * 2 * ... * 2 (n copies of 2)

[hexed barn]

which leads back to this

[balmy cypress]

Ok but for example I could have {a} and {a,b}, aren't there 2 choices for a here?

[hexed barn]

these are two different sets

what's the point going on here

note that you have not specified what A is

[lyric charm]

you misunderstand

[balmy cypress]

If A for example has 3 elements

[balmy cypress]

Yes

[lyric charm]

be concrete and tell us what EXACTLY you want to look at for A

A = {a,b,c}?

[balmy cypress]

Yes

[hexed barn]

then {a} and {a, b} correspond to {include, exclude, exclude} and {include, include, exclude} respectively

[lyric charm]

ok, then the subset {a} is represented by

Take a? YES
Take b? NO
Take c? NO

and {a,b} is represented by

Take a? YES
Take b? YES
Take c? NO

[balmy cypress]

Yes

Ok

And on what basis does 2* 2*...*2 n times give me all the possible combinations?

[lyric charm]

do you know the multiplication rule in combinatorics

[balmy cypress]

Newton ?

Binomial

[hexed barn]

binomial is going too far

[balmy cypress]

Combinations?

[hexed barn]

closer, but can you state the rule?

[balmy cypress]

Since I can choose each element 2 times all combinations are given by 2^n I think

Oh yes

[hexed barn]

you are not choosing each element up to 2 times. you are given 2 choices for each element

[balmy cypress]

Like a=0 off and a=1 on

So 2 options

[hexed barn]

correct, that is a better phrasing.

[balmy cypress]

And then n times to have all those of A

[hexed barn]

to consider each element of A*

but yes

[balmy cypress]

Clear

Thanks all

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[final girder]

if the coil is perpendicular to the magnetic field it's emf is 0 right?

[final girder]

because on the answer key it says maximum but what I learned it's 0

[blissful ether]

yes

[final girder]

because my answer sheet it says maximum

and my friend said maximum

[last slate]

You're talking about flemings right hand rule and laws on electromagnetic induction right?

[final girder]

idk I dont memorise the name but it's like change in flux and emf

[last slate]

Yeah that is electromagnetic induction definition

[final girder]

so is it 0 when perpendicular the emf or maximum

[last slate]

no, rather maximum

You want a visual representation?

The visual representation will be on A.C generator sin wave

[final girder]

but isn't flux maximum because cosin 0 is 1

and if flux is max then emf is 0

[last slate]

No no it's not like that, it is rate of change of magnetic flux

[final girder]

what

[last slate]

wait lemme show you the textbook version

[final girder]

this is what my teacher told me

[last slate]

Ys that exactly, look at the A.C graph when the coil rotates.
When the coil is perpendicular to the magnetic field , it is maximum and is the amplitude.
This wave occurs due to the slip rings of an A.C generator which reverse the current in armature coil
Thereby making this sine curve

[final girder]

yes the flux is maximum

but the emf is zero

[last slate]

Yes

[final girder]

I found a video it also says emf is zero but I'm confused cuz 3 people said its max and my answer key

[last slate]

No EMF is not zero, it is the highest peak of the curve

[last slate]

If EMF was zero, how can the coil turn.
There's a misunderstanding

Do you want me to guide through the working of the AC generator

[final girder]

look

[last slate]

Yep so emf according to that graph is 10 millivolts

[final girder]

it saisays time equals to 0 parallel and it's the max point on the graph

and it's 0 when perpendicular in the graph

[last slate]

The first option is wrong because at t = 0, coil must be perpendicular

[final girder]

chat gpt says its 0

my tea her said its 0

my quiz said its 0

but my course revision said its maximum

3 other people said maximum

[last slate]

Don't follow people, follow the concept what it says

donot follow gpt, it can make mistakes

Can you visualize the coil rotating?

[final girder]

I'll just try to look in my textbook to check thanks for helping

[final girder]

yes

[last slate]

I mean did you get the point?

[final girder]

no but I'll like try to understand

[last slate]

seems like you're confused

you can check the textbook and then be assured

[final girder]

do you use the angle theta for the flux equation as the surface to the field or the normal of coil to the field

[last slate]

You can use this formula wait

it might be useful later

[final girder]

which

[last slate]

yeah wait a min

e=e_0 sin⁡2πnt
This is the standard formula
e0 is the max emf induced, sin 2 x pi x n x t is complex but just remember sin theta

[final girder]

what's n

is e current

[last slate]

I said it is complex, did you learn this equation before

[final girder]

the one I did is I=Imaxsin(s pi frequency time)

[last slate]

that is maximum current induced

[final girder]

yes I did learn that

[last slate]

Oh great then you might be knowing it

[final girder]

but what does that have to do with the flux

[last slate]

this equation is in my textbook and I haven't learned it since it is complex and out of syllabi

[last slate]

Well that is the Law of electromagnetic induction by Faraday
Whenever there is a change in magnetic flux linked with a coil. an e.m.f is induced

Now I cannot explain why that happens, because it is a natural phenomena

[final girder]

I found another one that says its 0 when it's perpendicular

[last slate]

sin 90 is 1, how can emf be 0
Do not follow those

[final girder]

I use the rule flux is equals to magnetic field times area time cosin theta

where theta is the angle between the normal to the surface and the field

[last slate]

Look I think we are not coming to a conclusion as you're having misunderstandings here
You can check the textbook definition and check back later

[final girder]

I'll just watch videos thanks for explaining

how to close chat

[last slate]

Yeah that might help

Do you want a AC generator visualize video?

[final girder]

no its ok I'll find it

[last slate]

. close

[final girder]

.close

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[last slate]

Alright have a nice day man

[final girder]

you too

[surreal charm]

that is right

[midnight plank]

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[coral belfry]

,tex
Hi there. For a sequence $b_n $ that converges to a non-zero real number b, i wanna prove that $\frac{1}{b_n} $ converges to $\frac{1}{b} $. Is this proof gud?

\vspace{0.15cm}
$\forall \epsilon >0 , \exists n_0 \in\mathbb{N} : n>n_0 \implies \lvert b_n -b\rvert <\epsilon $

\vspace{0.2cm}
We have that: $\lvert b_n\rvert > \frac{\lvert b\rvert }{2} , \forall n>n_0$

\vspace{0.12cm}
$\lvert \frac{1}{b_n} -\frac{1}{b} \rvert = \lvert \frac{b -b_n}{bb_n} \rvert \leq \frac{\lvert b_n-b \rvert }{\lvert b\rvert } \frac{2}{\lvert b\rvert } <\frac{2\epsilon }{b^2} $

\vspace{0.2cm}
So for $\epsilon ' = \frac{2\epsilon }{b^2} >0 : \ n>n_0 \implies \lim_{n\to\infty } \frac{1}{b_n} = \frac{1}{b} $

[grand pond]

fijokazż

[visual tiger]

$\lvert b_n\rvert > \frac{\lvert b\rvert }{2} , \forall n>n_0$
What's your exact justification for that?

[grand pond]

Rafilouyear2026

[coral belfry]

we can prove it using a different n_k but i just took the max of the two to skip the step

[visual tiger]

Also you seem to be using a lot of n_0's interchangeably even though they can differ from one property to another

[visual tiger]

ok but for which epsilon?

[coral belfry]

|b|\2

we take cases for what b is

[visual tiger]

ok, so that's a bit that's missing

[coral belfry]

we alr had it proven in the notes so i didnt write it

i just took the maximum n_k such that both inequalities are true and called it n_0

my main problem is with whether i can just name a new epsilon to equal 2ε/b² and still say that we prove convergence

in the notes they do smth different i dont rember well

,tex
What they did in the notes was take an $n_1 $ such that $$n>n_1 \implies \ \lvert b_n -b \rvert < \frac{b^2 \epsilon }{2} $$ and then they started doing what i did. so in the end they just have $\lvert \frac{1}{b_n} -b^{-1} \rvert < \epsilon $ for a convenient choice of $n_k $

[grand pond]

fijokazż

[coral belfry]

but i wonder if what i did is also fine

well anyway ill js do it the normal way if its in the exams

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[hoary hamlet]

when dealing with this kind of problem, do i need to rearrange for the "there exists" variable? or is it the other way around

[olive matrix]

I don't understand what you mean; you can't really rearrange any of them.

[midnight plank]

@hoary hamlet Has your question been resolved?

[torn compass]

pretty sure he means to convert a universal statement into existential one or the other way around

[lyric charm]

i dont think there is really any shortcut to thinking long and hard about these sorts of statements

[hoary hamlet]

so for example, for statement 3 they advised to rearrange in terms of y

the model answers

[hoary hamlet]

i meant the algebra sorry

bad wording from me

[olive matrix]

oh like solving the expression / equation given for one of the variables

sure, that could help

[hoary hamlet]

is there also methods or anything that could make my life easier

[midnight plank]

@hoary hamlet Has your question been resolved?

[thin glade]

not directly related to problem solving per se, but having a solid foundation in logic wouldnt hurt

[golden crescent]

these typa problems are mathematical logic

[thin glade]

https://richardhammack.github.io/BookOfProof/Main.pdf#page=46
chapter 2 of hammack's book of proof is good for this (i read the entire book before uni, its good for learning how to prove stuff)

beware that lots of things are in symbols tho

[hoary hamlet]

i see

thanks

[midnight plank]

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[coral belfry]

hi. i need help with understanding something important. so, i was told that for any sequence a_n, f(a_n) is also a sequence, and not a function. that does make sense, but just now, i found on some notes, that a_n are functions N->R, and so that would make f(a_n) a function, too. My question is: when taking the limit of f(a_n), do we interpret it as a function, or as a sequence? im asking bc the epsilon definition of the limit may vary

[nova yoke]

you are correct, sequences are functions (with domain equal to the natural numbers)

[coral belfry]

alright. so lim_{n->infty} (f(a_n)) = l, how would we write that in a formal definition?

[main current]

There is a different notion of limit for functions R -> R, which you might be getting confused with. Note functions R->R are different from functions that might give a sequence

[coral belfry]

i see

[nova yoke]

but it wouldn't make much sense to use epsilon/delta for them, because the "delta" is used when you make the "input" to the function approach some fixed number, which is not possible when the input values are constrained to be integers

so like $\lim_{n \to 1} a_n$ wouldn't make sense

[grand pond]

Bungo

[coral belfry]

hmm alright

thanks for clarifying

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[gray wave]

I didn't understand, can someone explain it clearly?

[gray wave]

I need an explanation like a explained to a donkey

[midnight plank]

@gray wave Has your question been resolved?

[wet wagon]

Is there a specific part you want to have explained or do you want a walkthrough of everything

[midnight plank]

@gray wave Has your question been resolved?

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[gray wave]

everything

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[tall lily]

it's not divergent. what's wrong with my logic?

[tall lily]

last phrase is some point A between 0 and 1 if that's not clear

[sharp coral]

you are misusing the 1/x^p criterion

the criterion ``$p \le 1$ means divergent'' is for integrals of the form $$\int_1^\infty \frac{dx}{x^p}$$

[grand pond]

cloud ☁

[sharp coral]

integrals from 0 to 1 have essentially the opposite convergence criterion: p >= 1 means divergent, p < 1 means convergent (you can prove this)

[tall lily]

ah i see

thanks

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[cerulean oyster]

which do you think it is

[formal rune]

hm

[formal rune]

lowk no idea

[cerulean oyster]

Can you identify the equation that divides the two regions?

[formal rune]

yeah no

[cerulean oyster]

I mean, the purple line

It looks like a linear equation

[formal rune]

do i gotta find rise/run

[cerulean oyster]

Not really.

Tbh you should probably go over basic functions / equations if you cant recognize this one

but its y = -x

[formal rune]

any yt vid u recommne

[cerulean oyster]

no idea tbh

[quick creek]

place three dots in the blue area in different quadrants

and see all of the 3 satisfies the inequality

[formal rune]

what

[quick creek]

so for example pick a dot in the blue area

[formal rune]

i understand that y≥x means shade up and y≤x shade down

but here x is the first varaible

so is it the opposite

[quick creek]

<@&268886789983436800> scam

[formal rune]

lmao

so

[quick creek]

move the - to the left for the bottom two

then all four inequality is for y

[formal rune]

but x is the first variable

understand that y≥x means shade up and y≤x shade down
but here x is the first varaible

[quick creek]

change to x<=y
and x>=y

the first one is shade up
the second is shade down

[formal rune]

so so its basically the opposite